A question about common emitter amplifier

In summary, the conversation discusses the use of a blocking capacitor in a common emitter amplifier and the relationship between the capacitor and the input base bias resistors in order to function as a high pass filter. The equation C>= 1/2пf(R1||R2) is mentioned as a guide for choosing the appropriate value for the capacitor. The two main functions of the blocking capacitor are also mentioned as blocking DC and serving as a high pass filter.
  • #1
Harrison G
41
2

Homework Statement


Hello my fellow PF mates! Today i received my electronic textbook from Russia. The book is great, however i can't explain myself one single equation, but i hope you will help me:-)

The question is about the blocking capacitor in a common emitter amplifier. R1 and R2 are the voltage divider resistors (sorry, but i still haven't learned how to post images) and C is the blocking capacitor.
There is one relationship that i can't figure out.

Homework Equations


The textbook says that when choosing C, C>= 1/2пf(R1||R2)
My question is why C should be equal or greater then 1/2пf(R1||R2)?

The Attempt at a Solution

 
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  • #2
What are the major contributing factors to the input impedance of the circuit?
 
  • #3
gneill said:
What are the major contributing factors to the input impedance of the circuit?
The input impedance is 8kOhms and the frequency is 200Hz, for this is used a capacitor of 0.1uF
 
  • #4
Harrison G said:

Homework Statement


Hello my fellow PF mates! Today i received my electronic textbook from Russia. The book is great, however i can't explain myself one single equation, but i hope you will help me:-)

The question is about the blocking capacitor in a common emitter amplifier. R1 and R2 are the voltage divider resistors (sorry, but i still haven't learned how to post images) and C is the blocking capacitor.
There is one relationship that i can't figure out.

Homework Equations


The textbook says that when choosing C, C>= 1/2пf(R1||R2)
My question is why C should be equal or greater then 1/2пf(R1||R2)?

The Attempt at a Solution

What are the two functions of the input DC blocking capacitor in the CE amplifier circuit?

http://webpages.ursinus.edu/lriley/ref/circuits/img119.gif
img119.gif
 
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  • #5
berkeman said:
What are the two functions of the input DC blocking capacitor in the CE amplifier circuit?

http://webpages.ursinus.edu/lriley/ref/circuits/img119.gif
img119.gif
As far as i know, the blocking capacitor have 2 main uses. 1st they block DC and allow only AC to pass and 2nd they incidentaly serve as a high pass filter and must be calculated for the working frequency so that they don't cause any trouble. But as i see now, maybe i didnt formulated correctly my question -- How do we calculate that capacitor and what its value depends upon?
 
  • #6
Harrison G said:
As far as i know, the blocking capacitor have 2 main uses. 1st they block DC and allow only AC to pass and 2nd they incidentaly serve as a high pass filter and must be calculated for the working frequency so that they don't cause any trouble. But as i see now, maybe i didnt formulated correctly my question:How do we calculate that capacitor and what its value depends upon?
Correct. So to be a HPF, that input C and the two input base bias resistors have to have a certain relationship. And it looks like that relationship is the equation you posted in your first post... :smile:
 
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  • #7
berkeman said:
Correct. So to be a HPF, that input C and the two input base bias resistors have to have a certain relationship. And it looks like that relationship is the equation you posted in your first post... :smile:
Thank you verry much for youre help! :-)
 
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Related to A question about common emitter amplifier

1. What is a common emitter amplifier?

A common emitter amplifier is a type of electronic circuit used to amplify a small signal to a larger one. It is a three-terminal device, with the input signal applied to the base terminal, the output taken from the collector terminal, and the emitter terminal connected to ground. It is commonly used in audio amplifiers, radio receivers, and other electronic devices.

2. How does a common emitter amplifier work?

A common emitter amplifier works by using a transistor to amplify the input signal. As the input signal is applied to the base terminal, it causes a small current to flow between the base and emitter. This current then controls a larger current between the collector and emitter, resulting in a larger output signal. The amplification is achieved through the transistor's ability to control the flow of current between its terminals.

3. What are the advantages of using a common emitter amplifier?

One of the main advantages of using a common emitter amplifier is its high gain, meaning it can amplify small input signals to larger outputs. It also has a relatively simple circuit design and can be easily biased for linear operation. Additionally, it has a wide bandwidth, making it suitable for amplifying a range of frequencies.

4. What are the limitations of a common emitter amplifier?

One limitation of a common emitter amplifier is its relatively low input impedance, which can lead to signal loss and distortion. It also has a relatively high output impedance, which can limit its ability to drive low impedance loads. Additionally, it is susceptible to temperature variations and noise interference.

5. How is the gain of a common emitter amplifier calculated?

The gain of a common emitter amplifier can be calculated by dividing the change in output voltage by the change in input voltage. This is known as the voltage gain and is typically expressed as a ratio or in decibels (dB). The gain can also be calculated using the ratio of output current to input current, known as the current gain. The value of the gain is dependent on the circuit design and the biasing of the transistor.

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